Why vertex and edge transitivity on a k-regular nonsymmetric graph implies even k

Question

A graph is half transitive if it's both vertex and edge transitive but not arc transitive... Also it must be k-regular...

The question is why every half transitive graph is even?

(there is no trivial example)

The solution seems to be simple because it's an elementary HW.

Thanks...

This question was asked but no answer was given and Im tired of trying to solve it. Can anyone at least give a hint or an answer? I'm a self learner and have tried looking it up but all the references are garbage and dont show proofs.

Answer 1

HINT: If a graph is edge-transitive but not arc-transitive, the set $A$ of arcs of $G$ are partitioned by the automorphism group of $G$ into precisely two orbits $A_1$ and $A_2$, where $uv \in A_i \Leftrightarrow vu \in A_2$. Note that every vertex $v$ is incident to the same number $k$ of arcs in $A_1$ as $A_2$ where $k$ is the degree of $G$, and for each $i=1,2$, all vertices have the same degree in $A_i$ which implies that each vertex has both indegree and outdegree 0 in $A_i$.

Can you finish from here?